Articulation Points in Graph in DSA Java - Time & Space Complexity

Time Complexity: Articulation Points in Graph

O(n + m)

Understanding Time Complexity

We want to understand how the time needed to find articulation points changes as the graph grows.

How does the algorithm's work increase when the graph has more nodes and edges?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void dfs(int u, int parent) {
    visited[u] = true;
    disc[u] = low[u] = ++time;
    int children = 0;
    for (int v : graph[u]) {
        if (!visited[v]) {
            children++;
            dfs(v, u);
            low[u] = Math.min(low[u], low[v]);
            if (parent == -1 && children > 1) articulation[u] = true;
            if (parent != -1 && low[v] >= disc[u]) articulation[u] = true;
        } else if (v != parent) {
            low[u] = Math.min(low[u], disc[v]);
        }
    }
}

This code finds articulation points using DFS, tracking discovery and low times for each node.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Depth-first search (DFS) visits each node and its edges.
• How many times: Each node is visited once, and each edge is checked once or twice.

How Execution Grows With Input

As the graph grows, the algorithm visits all nodes and edges once.

Input Size (n nodes, m edges)	Approx. Operations
10 nodes, 15 edges	About 25 operations
100 nodes, 200 edges	About 300 operations
1000 nodes, 5000 edges	About 6000 operations

Pattern observation: Operations grow roughly in proportion to nodes plus edges.

Final Time Complexity

Time Complexity: O(n + m)

This means the time grows linearly with the number of nodes and edges in the graph.

Common Mistake

[X] Wrong: "The algorithm checks every possible path, so it must be exponential."

[OK] Correct: The DFS visits each node and edge only once, avoiding repeated work, so it runs in linear time.

Interview Connect

Understanding this linear time complexity helps you explain how graph traversal algorithms efficiently find key points in networks.

Self-Check

"What if the graph is disconnected and we run DFS from every node? How would the time complexity change?"

Practice

1. What is an articulation point in a graph?

easy

A. A vertex that forms a cycle in the graph

B. A vertex with the highest degree in the graph

C. A vertex connected to all other vertices

D. A vertex which, when removed, increases the number of connected components

Solution

1. Step 1: Understand the definition of articulation point

   An articulation point is a vertex that, if removed, breaks the graph into more pieces (connected components).

2. Step 2: Compare options with definition

   Only A vertex which, when removed, increases the number of connected components matches this definition exactly.

3. Final Answer:

   A vertex which, when removed, increases the number of connected components -> Option D

4. Quick Check:

   Articulation point = vertex breaking connectivity [OK]

Hint: Articulation point breaks graph connectivity when removed [OK]

Common Mistakes:

• Confusing articulation point with highest degree vertex
• Thinking articulation point forms cycles
• Assuming articulation point connects to all vertices

2. Which of the following is the correct way to declare a graph adjacency list in Java for articulation point detection?

easy

A. int graph[][] = new int[];

B. HashMap graph = new HashMap<>();

C. ArrayList graph = new ArrayList<>();

D. List<Integer> graph = new List<>();

Solution

1. Step 1: Understand adjacency list structure

   An adjacency list for a graph is usually a list of lists, where each inner list holds neighbors of a vertex.

2. Step 2: Check Java syntax for adjacency list

   ArrayList graph = new ArrayList<>(); correctly declares an ArrayList of ArrayLists of Integers, suitable for adjacency list.

3. Final Answer:

   ArrayList graph = new ArrayList<>(); -> Option C

4. Quick Check:

   Adjacency list = list of lists [OK]

Hint: Adjacency list uses list of lists in Java [OK]

Common Mistakes:

• Using 2D array without size initialization
• Using HashMap with wrong value type
• Trying to instantiate interface List directly

3. Consider the following Java code snippet for articulation point detection. What will be the output when the graph has edges: 0-1, 1-2, 1-3?

int V = 4;
ArrayList graph = new ArrayList<>();
for (int i = 0; i < V; i++) graph.add(new ArrayList<>());
graph.get(0).add(1);
graph.get(1).add(0);
graph.get(1).add(2);
graph.get(2).add(1);
graph.get(1).add(3);
graph.get(3).add(1);
// Articulation points found: 1

medium

A. []

B. [1]

C. [2, 3]

D. [0, 1]

Solution

1. Step 1: Analyze the graph structure

   The graph has vertices 0,1,2,3 with edges 0-1, 1-2, 1-3. Vertex 1 connects to all others.

2. Step 2: Identify articulation points

   Removing vertex 1 disconnects the graph into three parts, so 1 is an articulation point. Others are not.

3. Final Answer:

   [1] -> Option B

4. Quick Check:

   Only vertex 1 breaks connectivity [OK]

Hint: Check which vertex removal splits graph [OK]

Common Mistakes:

• Assuming leaf nodes are articulation points
• Including vertices with degree 1 as articulation points
• Ignoring connectivity after removal

4. The following Java code snippet is intended to find articulation points but has a bug. What is the error?

void dfs(int u, int parent) {
  visited[u] = true;
  disc[u] = low[u] = time++;
  int children = 0;
  for (int v : graph.get(u)) {
    if (!visited[v]) {
      children++;
      dfs(v, u);
      low[u] = Math.min(low[u], low[v]);
      if (parent != -1 && low[v] >= disc[u])
        articulation[u] = true;
    } else if (v != parent) {
      low[u] = Math.min(low[u], disc[v]);
    }
  }
  if (parent == -1 && children > 1)
    articulation[u] = true;
}

medium

A. The time variable is not initialized before DFS

B. The condition for root articulation point is incorrect

C. The low array is not updated correctly for back edges

D. The visited array is not marked true for vertex u

Solution

1. Step 1: Check initialization of time variable

   The variable 'time' is used to assign discovery times but must be initialized before DFS starts.

2. Step 2: Verify other parts of DFS

   Conditions for root and back edges are correct; visited[u] is marked true properly.

3. Final Answer:

   The time variable is not initialized before DFS -> Option A

4. Quick Check:

   Uninitialized time causes wrong discovery times [OK]

Hint: Initialize time counter before DFS call [OK]

Common Mistakes:

• Forgetting to initialize time before DFS
• Misusing parent check for root node
• Not updating low values for back edges

5. Given a connected undirected graph with 5 vertices and edges: 0-1, 1-2, 2-0, 1-3, 3-4, which vertices are articulation points?

hard

A. [1, 3]

B. [0, 2]

C. [3]

D. [1]

Solution

1. Step 1: Analyze the graph structure

   Vertices 0,1,2 form a triangle (cycle). Edges 1-3 and 3-4 form a chain attached to vertex 1.

2. Step 2: Identify articulation points

   Removing vertex 1 disconnects 3 and 4 from the cycle, so 1 is an articulation point. Removing 3 disconnects 4, so 3 is also an articulation point. Vertices 0 and 2 are in a cycle, so not articulation points.

3. Final Answer:

   [1, 3] -> Option A

4. Quick Check:

   Vertices breaking connectivity: 1 and 3 [OK]

Hint: Check vertices whose removal splits graph into parts [OK]

Common Mistakes:

• Assuming cycle vertices are articulation points
• Missing vertex 3 as articulation point
• Only considering vertex 1 as articulation point